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# □ show a/√(a² + 8bc) + b/√(b² + 8ca) + c/√(c² + 8ab) ≥ 1 .?

a,b,c > 0.

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- IndicaLv 72 years agoFavorite Answer
This is easy by Jensen using the convex function 1/√x (x>0)

LHS is homogeneous so you can assume that a+b+c=1 and then apply Jensen with weights a,b,c for

∑a/√(a²+8bc) ≥ 1/(√∑a(a²+8bc)) = 1/√(∑a³+24abc) … (i)

But (∑a)³ = (a+b+c)³ = ∑a³ + 3∑a²(b+c) + 6abc ≥ ∑a³ + 3(6 ⁶√(a⁶b⁶c⁶)) + 6abc by AM-GM

Hence 1 = (∑a)³ ≥ ∑a³ + 24abc and (i) then gives ∑a/√(a²+8bc) ≥ 1/1 = 1

Equality occurs when a=b=c

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